Could you please comment on the legibility of my proof of the Rank-Plus-Nullity Theorem?
Theorem: $A$ is an $(m \times n)$-matrix with $k$ pivot columns. The dimension of the null space of $A$ is $n - k$.
The linear algebra books to which I referred only stated it in the context of linear transformations. So, the verifications of it relied on various propositions about arbitrary vector spaces, but I wanted a demonstration for it in the context of matrices. Here is my demonstration:
$T$ denotes the set of every integer $1 \leq j \leq n$ such that the $j^{\mathrm{th}}$ column vector in $A$ is not a pivot column. $\vert T \vert = n - k$. For each integer $j \in T$, $\boldsymbol{x}_{j} = (x_{i})_{\scriptscriptstyle{i=1}}^{\scriptscriptstyle{n}}$ is the vector in the null space of $A$ defined so that $x_{j} = 1$ and $x_{i} = 0$ for every integer $i \in T\setminus\{j\}$. $X = \{\boldsymbol{x}_{j} \mid j \in T\}$ is a linearly independent set in the null space of $A$. If $X$ spans the null space of $A$, $X$ is a basis for the null space of $A$.
$R$ is the reduced-row echelon matrix that is row equivalent to $A$. The null space of $R$ is the same as the null space as $A$. For any vector $\boldsymbol{y} = (y_{i})_{\scriptscriptstyle{i=1}}^{\scriptscriptstyle{n}}$ in the null space of $A$, \begin{equation*} \boldsymbol{y}^{\prime} = y_{t_{1}}\boldsymbol{x}_{t_{1}} + \ldots + y_{t_{n-k}}\boldsymbol{x}_{t_{n-k}} . \end{equation*} Being a linear combination of vectors in the null space of $A$, $\boldsymbol{y}^{\prime}$ is a vector in the null space of $A$. \begin{equation*} \boldsymbol{y} - \boldsymbol{y}^{\prime} = (z_{i})_{\scriptscriptstyle{i=1}}^{\scriptscriptstyle{n}} \end{equation*} is another vector in the null space of $A$. Moreover, $z_{j} = 0$ for each integer $j \in T$. Since for every integer $1 \leq i \leq n$ that is not in $T$, $z_{i}$ is equal to a linear combination from $\{z_{j} \mid j \in T\}$, $z_{i} = 0$, too. Consequently, $\boldsymbol{y} - \boldsymbol{y}^{\prime} = \boldsymbol{0}_{\scriptscriptstyle{{\mathbb{R}}^{m}}}$, or equivalently, $\boldsymbol{y} = \boldsymbol{y}^{\prime}$. $X$ spans the null space of $A$.